Traducciones

Let {ui,M2,M3} and {vi,V2,V3} be bases for U and V (respectively). Then, the set {mi,M2,M3,
'Vi,V2,V3] is linearly dependent, since (acronymref | theorem | G) says we cannot have 6 linearly
independent vectors in a vector space of dimension 5. So we can assert that there is a non-trivial
relation of linear dependence,

ai«i -I- 02^2 + cisus + bivi + b2V2 + bsvs =

where ai, a2- as and foi, 62, &3 are not all zero. We can rearrange this equation as

aiui + a2U2 + aaus = - bivi - b2V2 - b^v^

This is an equality of two vectors, so we can give this common vector a name, say w,

w = aiUi + a2M2 + aaMs = - bivi - b2V2 - b^vz

This is the desired non-zero vector, as we will now show. First, since w = ami + 02^2 + as^s , we
can see that w £U. Similarly, w = — b^vi — 621^2 — bsvz, so w € F. This establishes that
w £U nV ({acronymref| definition I SI)). Is w ^ Ql Suppose not. in other words, suppose w = Q.
Then

= w = ai«i -I- 02^2 + asMs

Because {tti, U2, ws} is a basis for [/, it is a linearly independent set and the relation of linear
dependence above means we must conclude that ai = 02 = ^3 = 0. By a similar process, we would
conclude that 61 = 62 = ^a = 0. But this is a contradiction since ai, a2, a3,bi,b2, 63 were chosen so
that some were nonzero. So w ^^^ 0. How does this generalize? All we really needed was the orig-
inal relation of linear dependence that resulted because we had "too many" vectors in W. A
more general statement would be: Suppose that w is a vector space with dimension n, U is a
subspace of dimension p and V is a subspace of dimension q. li p+ q>n, then U CiV contains a
non-zero vector.

Sean {ui, U2. M3} y {vi, V2, v^] bases para UyV (respectivamente). Luego, el conjunto {iti,
■"2, M3, i^i, V2, V3} es linealmente independiente, ya que (acronymref | theorem | G) dice que no
podemos tener 6 vectores linealmente independientes en un espacio vectorial de dimension 5.
For lo tanto, podemos afirmar que existe una relacion no trivial de dependencia lineal,

aiUi + a2U2 + a^uz + bivi -\- b2V2 + 631^3 =

donde 01,02,03 y 6i,&2,&3 no son todos ceros. Fodemos cambiar esta ecuacion como

oiMi -I- 02^2 + a^uz = - bivi - b2V2 - bavz

Esto es una igualdad de dos vectores, por lo cual podemos darle al vector comiin un nombre, Ha-
ni ado IJO

W = OiMi -I- 02^2 + 03^3 = - blVi - b2V2 - bzVz

Este es el vector deseado diferente de cero, como se muestran ahora. Frimero, desde w = a\ui +
02^2 + asuz, podemos ver que w gU. Igualmente w = — bivi — b2V2 — bzvz, de modo que w gV.
Esto establece que w £U CiV ((acronymref| definition! SI)). ^Es w^Ol Suponga que no, en otras
palabras, suponga w = 0. Luego

= w = oiMi -I- a2M2 + azuz

Dado que {mi. M2, ms} es una base de U, es un conjunto linealmente independiente y la relacion
de dependencia lineal anterior significa que debemos concluir que fei = 62 = ^3 = 0. Pero es una
contradiccion desde ai, 02, 03, 61, 62, ^3 fueron elegidos tal que fueran diferentes de cero. Asi
10 :^ 0. ^Como se puede generalizar? Todo lo que realmente se necesitaba era la original relacion
de dependencia lineal que dio lugar, ya que habia "demasiados" vectores en W. Una declaracion
mas general seria: Suponga que w es un espacio vectorial de dimension n, U es un subespacio de
dimension p y 1/ es un subespacio de dimension q. Si p + q> n. luego U CiV contiene un vector
diferente de cero.

Contributed by Robert Beezer
Contribuido por Robert Beezer
Traducido por Felipe Pinzon